Calculus: Early Transcendentals, Chapter 6, 6.2, Section 6.2, Problem 13

y=1+sec(x),y=3
Refer the image. From the graph, the curves intersects at x=-pi/3 and x=pi/3.
Using washer method,
A cross section is a washer of cross sectional area A(x) with,
Inner radius=(1+sec(x))-1=sec(x)
Outer radius=3-1=2
A(x)=pi(2^2-(sec(x))^2)
A(x)=pi(4-sec^2(x))
Volume of the solid obtained by rotating the region bounded by the given curves about y=1 (V) is,
V=int_(-pi/3)^(pi/3)A(x)dx
V=int_(-pi/3)^(pi/3)pi(4-sec^2(x))dx
V=2piint_0^(pi/3)(4-sec^2(x))dx
V=2pi[4x-tan(x)]_0^(pi/3)
V=2pi((4*pi/3-tan(pi/3))-(4*0-tan(0)))
V=2pi(4*pi/3-sqrt(3)-0)
V=2pi((4pi)/3-sqrt(3))